Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
long long L[N], R[N];
long long sum[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long s, a, b;
  cin >> s >> a >> b;
  s *= -1;
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> L[i] >> R[i];
    sum[i] = R[i] - L[i] + sum[i - 1];
  }
  int q;
  cin >> q;
  while (q--) {
    long long x, y;
    cin >> x >> y;
    double ans = 0;
    double cnt = (double)s / y + 1;
    int la = 0, ra = n;
    while (la < ra) {
      int m = (la + ra + 1) >> 1;
      if (s * (L[m] - x) + L[m] * y < a * y) {
        la = m;
      } else
        ra = m - 1;
    }
    if (la > 0) ans += max(0.0, (double)(s * (R[la] - x)) / y + R[la] - a);
    ans -= sum[la] * cnt;
    int last = la;
    la = 1, ra = n + 1;
    while (la < ra) {
      int m = (la + ra) >> 1;
      if (s * (R[m] - x) + R[m] * y > b * y) {
        ra = m;
      } else
        la = m + 1;
    }
    if (la <= n) ans += max(0.0, b - (double)(s * (L[la] - x)) / y - L[la]);
    ans += sum[la - 1] * cnt;
    if (last == la) ans = b - a;
    cout << setprecision(15) << fixed << ans << "\n";
  }
  return 0;
}